Question

Laplace transform of

\[g(t) =t^2\sinh(3t)\]
\[G(p) =\mathcal L\big\{t^2\sinh(3t)\big\}\]

Answer 1

is it best to use \[\mathcal L\{tf(t)\}=-\frac{d}{dx}\mathcal L\{f(t)\}\] a couple of times?

Answer 2

and then the quotient rule?

Answer 3

Or instead to substitute in the exponential identity for the hyperbolic trig function

Answer 4

Does this approach (see attached) make sense to you?

Answer 5

i think the solution maybe (1/((p-3)^3)-(1/((p+3)^3)

Answer 6

yeah that method works

Answer 7

is there some way to simplify the numerator?

Answer 8

I have not tried, too tedious

Answer 9

i had this working

Answer 10

\[18(s^2+3)=(s+3)^3-(s-3^3)\] ?

Answer 11

I think best expression is:\[\frac{ 1 }{ (s-3)^3 }-\frac{ 1 }{ (s+3)^3 }\]

Answer 12

@unkleRhaukus:ya...i also followed the same method.

Answer 13

how can i get my answer to equal that nice expression ?

Answer 14

if u use partial fractions..u can get the expression got by carlosGP!

Answer 15

ill try that now @susanka

Answer 16

i got

\[\frac{p^2+3}{(p^2-9)^3}=\frac1{(p^2-9)^2}+\frac3{(p^2-9)^3}\]

Answer 17

gr8:)

Answer 18

I also get the same as @CarlosGP ,
I substitute into exponential identity for the hyperbolic function

Answer 19

@UnkleRhaukus
You must solve:
\[\frac{ 18p^2+54 }{ (p^2-9)^3 }=\frac{ A }{ (p-3) }+\frac{ B }{ (p-3)^2 }+\frac{C }{ (p-3)^3 }+\frac{ D }{ (p+3) }+\frac{ E }{ (p+3)^2 }+\frac{ F }{ (p+3)^3 }\]To get C=1 and F=-1 (A=B=D=E=0)

Answer 20

oh, i though it was
\[\frac{p^2+3}{(p^2-9)^3}=\frac{Ap+B}{p^2-9}+\frac{Cp+D}{(p^2-9)^2}+\frac{Ep+F}{(p^2-9)^3}\]

Answer 21

ya..but carlosGP is right..

Answer 22

@UnkleRhaukus with your formula, what happens with the p^2-9 of the denominator? it can also be factorized

Answer 23

im not very good with cubics

Answer 24

it seems it's easier by substitute the hyperbolic function by exponential function ... :D

Answer 25

Never mind! You made me hesitate. That means I was not very convinced either

Answer 26

where'd you get that partial fractions substitution @CarlosGP
it doesn't match with my table

Answer 27

Second case

Answer 28

but there is a p^2 in the denominator

Answer 29

True, but still can be factorized:\[(p^2-9)^3=(p-3)^3(p+3)^3\] You have applied fourth case that like third case is applied when polynomial ax^2+bx+c has no real solutions and cannot be factorized. In our case we have a polynomial with roots 3 and -3 with multiplicity =3